# Proof Without The Basis Extension Theorem

I need to prove the following theorem:

Let $$V$$ be a vector space over $$F$$. Let $$v_1,v_2,...,v_r \in V$$ and $$r > dim(V)$$. Then, $$(v_1,v_2,...,v_r)$$ is linearly dependent.

My Proof Attempt:

Let dim(V) = n, for convenience, and let $$(u_1,u_2,...,u_n)$$ be a basis for V. Clearly, $$(v_1,v_2,....,v_r)$$ cannot be a basis for V since r>n and all bases have the same length.

Now, all the vectors in $$(v_1,v_2,...,v_r)$$ can be written as linear combinations of the basis vectors, which generate V. Hence:

$$L(v_1,v_2,...,v_r) = L(u_1,u_2,....,u_n) = V$$

Since $$(v_1,v_2,...,v_r)$$ generates V but is not a basis, it follows that it cannot be linear independent. Hence, it is linearly dependent.

The book gives a proof that is based off of the Basis Extension Theorem and I understand that proof. I was just wondering if my approach is valid or not, since I attempted to prove the result before I looked at the one given by the book.

Your equality $$L(v_1,v_2,...,v_r) = L(u_1,u_2,....,u_n)$$ is not true. The point is that $$v_i$$ is a linear combination of the $$e_j's$$ , so you have $$\subset$$, but there is no guarantee that any $$e_j$$ may be written as a linear combination of the $$v_i's$$, because there is no reason why $$(v_1,\ldots,v_r)$$ should be a generating family of $$V$$.
Think about the case $$v_1=v_2=\cdots=v_r$$ and $$\dim_F(V)\geq 2$$ for example.
• Hmm but that seems to be a non-issue though? If $v_1 = v_2 = v_3 = .... = v_r$, then you're really just considering a list of length 1 and the dimension of your vector space is just 0, by virtue of the condition that r>n? – Abhi Feb 13 '20 at 16:01
• No. I can use repetition, since you are talking about families of vectors. If you prefer, you could take $v_1=v, v_2=2v,\ldots,v_r=rv$ for some non zero $v$. Anyway, your argument is incorrect, for the reason i mentioned. – GreginGre Feb 13 '20 at 16:07